Additional Mathematics 4037/11 — May/June 2018

Walkthrough

The system pairs a straight line with a circle, so we expect two intersection points. The linear equation $y - x = 4$ is rearranged to give $y = x + 4$, which is then substituted everywhere in the circle equation so that only one variable remains.

Expanding $(x+4)^2 = x^2 + 8x + 16$ and distributing $-4(x+4) = -4x - 16$, most terms cancel nicely, leaving $2x^2 - 4x - 16 = 0$. Dividing by 2 gives the simple quadratic $x^2 - 2x - 8 = 0$, which factorises as $(x-4)(x+2) = 0$, giving $x = 4$ or $x = -2$.

Each $x$ value is fed back through $y = x + 4$ to get its matching $y$: when $x = 4$, $y = 8$; when $x = -2$, $y = 2$. These are the two points where the line cuts the circle.

Key Takeaways

• To solve a linear–quadratic simultaneous system, substitute the linear equation into the quadratic to reduce it to one variable.
• Always pair each root with its corresponding value of the other variable — a solution is an ordered pair, not just an $x$ value.
• Simplifying by dividing out common factors before factorising makes the quadratic easier to handle.

Common Mistakes

• Forgetting to substitute into every occurrence of $y$ (both the $y^2$ term and the $-4y$ term).
• Sign slips when expanding $(x+4)^2$ — writing $x^2 + 16$ instead of $x^2 + 8x + 16$.
• Giving only the $x$ values without the matching $y$ values; the mark scheme awards one mark per correct pair.
• Stopping at $2x^2 - 4x - 16 = 0$ without solving it fully.

Things to Be Careful About

• Both solution pairs are required — there are two intersection points of a line with a circle here, and the mark scheme awards A1 for each pair.
• The mark scheme accepts either variable as the subject ($y = x + 4$ or $x = y - 4$), but the substitution must be shown and simplified to a three-term quadratic for the method marks.
• Check answers by substituting back into the original equations if time allows.